QJEEM25Ch9683: JEE Main 2025 Chemistry - Class 12 JEEM25Ch9683 Solution: Given below is the plot of the molar conductivity vs for KCl in …

$\begin{aligned}& \mathrm{R}=\rho \frac{\ell}{\mathrm{A}} \\& \kappa=G \cdot G^* \quad G=\frac{1}{\mathrm{R}} ; \kappa=\frac{1}{\rho} \\& \mathrm{G}^*=\frac{\ell}{\mathrm{A}} \\& \mathrm{R}=\text { Resistance } \\& \rho=\text { Resistivity } \\& \frac{\ell}{\mathrm{A}}=\text { cell } \operatorname{constant}\left(\mathrm{G}^*\right) \\& \frac{\kappa_{\mathrm{c}}}{\kappa_{\mathrm{d}}}=\frac{\mathrm{R}_{\mathrm{d}}}{\mathrm{R}_{\mathrm{c}}} ; \lambda_{\mathrm{m}}=\frac{\kappa \times 1000}{\mathrm{C}} \\& \frac{\kappa_{\mathrm{c}}}{\kappa_{\mathrm{d}}}=\frac{\left(\lambda_{\mathrm{m}} \cdot \mathrm{C}\right)}{\left(\lambda_{\mathrm{m}} \cdot C\right)_{\mathrm{d}}}=\frac{\mathrm{R}_{\mathrm{d}}}{R_{\mathrm{c}}} \quad \begin{array}{l}\mathrm{c}=\text { concentrated sol. } . \\\mathrm{d}=\text { diluted solution }\end{array} \\& \frac{100 \cdot(0.15)^2}{150 .(0.1)^2}=\frac{\mathrm{R}_{\mathrm{d}}}{100} \\& \mathrm{R}_{\mathrm{d}}=150 \Omega\end{aligned}$